orl wrote: A circle is tangential to sides $AB$ and $AC$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$. What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended? Assume that when correcting the typo, the problem would begin "A circle is tangential to sides $AB$ and $BC$..." The 1st circle $(P)$ tangent to the sides $AB$ and $AC$ is centered on the internal bisector of the angle $\angle ABC$. The points $E, F$ lie both inside of the segment $AC$. The 2nd circle $(Q)$ tangent to the lines $DA, DC$ has to be centered the internal bisector of the angle $\angle CDA$. In order to pass through the points $E, F$, it also has to be centered on the perpendicular bisector of the segment $EF$. Let $Q$ be the intersection of the the internal bisector of the angle $\angle ABC$ with the perpendicular bisector of the segment $Q$. Let $(Q)$ be a circle centered at this point and passing through the points $E, F$. The quadrilateral diagonal $AC \equiv EF$ is the radical axis of the circles $(P), (Q)$. Iff the circle $(Q)$ is tangent to the lines $DA, DC$ at points $K, L$, the tangent lengths $AG = AK$ to the circles $(P), (Q)$ from the vertex $A$ are equal and the tangent lengths $CH = CL$ to the circles $(P), (Q)$ from the vertex $C$ are also equal, because these 2 vertices are on the radical axis of the circles $(P), (Q)$. The tangent lengths $BG = BH$ to the circle $(P)$ from the vertex $B$ are equal the tangent lengths $DK = DL$ to the circle $(Q)$ from the vertex $D$ are also equal. Consequently, $AB + CD = (AG + BG) + (CL + DL) = AK + BH + CH + DK =$ $= (BH + CH) + (DK + AK) = BC + DA$ which is equivalent to the quadrilateral $ABCD$ being tangential. Since the circles $(P), (Q)$ are supposed to be different, the given circle $(P)$ must not be identical with the quadrilateral incircle $(I)$.

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